A point charge q is performing simple harmonic oscillations of amplitude A at
angular frequency \omega. Using Larmor's formula, the power radiated by the charge is
proportional to
Options:
(A) q \omega2A2
(B) q \omega4A2
(C) q2\omega2A2
(D) q2\omega4A2
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.16)**
**1. Electrodynamics & Radiation Principles**:
- **Larmor's Formula**: The instantaneous total power radiated by a non-relativistic point charge $q$ with acceleration $a(t)$ in vacuum is:
$P(t) = \frac{q^2 a(t)^2}{6\pi \epsilon_0 c^3}$
- The time-averaged power radiated is proportional to:
$\langle P \rangle \propto q^2 \langle a^2 \rangle$
**2. Detailed Step-by-Step Derivation**:
- The point charge undergoes simple harmonic oscillation of amplitude $A$ and angular frequency $\omega$:
$x(t) = A \cos(\omega t)$
- Differentiating twice to obtain acceleration:
$v(t) = \dot{x}(t) = -A\omega \sin(\omega t)$
$a(t) = \ddot{x}(t) = -A\omega^2 \cos(\omega t)$
- Squaring the acceleration:
$a(t)^2 = A^2 \omega^4 \cos^2(\omega t)$
- Taking the time-average over one complete period:
$\langle a^2 \rangle = A^2 \omega^4 \langle \cos^2(\omega t) \rangle = \frac{1}{2} A^2 \omega^4$
- Substituting into Larmor's relation:
$\langle P \rangle \propto q^2 \cdot (A^2 \omega^4) = q^2 \omega^4 A^2$
**3. Official Verification**:
The verified official IIT Master Answer is **Option (D)** (`q^2 \omega^4 A^2`).
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